anomalousdecay
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Maybe try thinking of "reversing" your thought process.
What do you think it could possibly converge to?
Might be easier to work from that.
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Nice probability question. (2 Viewers)
- Thread starter seanieg89
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What do you think it could possibly converge to?
Might be easier to work from that.
RealiseNothing
what is that?It is Cowpea
Nah my original working had a slight flaw (forgot to do 1-P(m)).
Re-doing my working gives:
Ain't gonna lie though, I'm really not confident with this answer.
anomalousdecay
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I would think for a question like this, the way you start is the most crucial factor in getting the end result.Nah my original working had a slight flaw (forgot to do 1-P(m)).
Re-doing my working gives:
Ain't gonna lie though, I'm really not confident with this answer.
Its because with probability you need to understand the case you are trying to address.
Though I can't even begin to start a question like this so I have no clue.
RealiseNothing
what is that?It is Cowpea
Anyway here's my working out/thought process. Some one might be able to see what's going on.
To find
we will find the amount of ways we can have a string of 'n' heads after exactly 'm' throws, then divide by the total amount of possibilities.
Now the total amount of possibilities is just
as there are 'm' throws, two outcomes.
Now for the other bit. The last n+1 throws must be T, H, H, ..., H, H to have a run of 'n' heads after 'm' throws. So this can only happen in one way, and thus we only need to find the amount of ways to arrange the first m-n-1 throws without having a string of 'n' heads.
To do this, we will find the amount of possibilities where we do have a string of 'n' heads, and subtract this from the total ways of arranging the first m-n-1 throws.
So assume we have a run of 'n' heads, and call this element A. Now we have to arrange the element A amongst m-2n-1 throws. Element A can be placed in (m-2n) positions, and the rest of the throws have two possibilities each - T or H. So we end up with the total ways being:
2^{m-2n-1})
The logic behind this is that it also contains the cases where we have a string of 'n+k' heads as we have a minimum of 'n' heads, and by "randomising" the remaining throws, it is possible for us to end up with a larger string of heads.
So the total ways where we don't get a string of 'n' heads is:
2^{m-2n-1})
Dividing this by the total ways gives:
2^{m-2n-1}}{2^m})
After some simplifying:
To find
Now the total amount of possibilities is just
Now for the other bit. The last n+1 throws must be T, H, H, ..., H, H to have a run of 'n' heads after 'm' throws. So this can only happen in one way, and thus we only need to find the amount of ways to arrange the first m-n-1 throws without having a string of 'n' heads.
To do this, we will find the amount of possibilities where we do have a string of 'n' heads, and subtract this from the total ways of arranging the first m-n-1 throws.
So assume we have a run of 'n' heads, and call this element A. Now we have to arrange the element A amongst m-2n-1 throws. Element A can be placed in (m-2n) positions, and the rest of the throws have two possibilities each - T or H. So we end up with the total ways being:
The logic behind this is that it also contains the cases where we have a string of 'n+k' heads as we have a minimum of 'n' heads, and by "randomising" the remaining throws, it is possible for us to end up with a larger string of heads.
So the total ways where we don't get a string of 'n' heads is:
Dividing this by the total ways gives:
After some simplifying:
seanieg89
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Without latex rendering properly atm, I won't look carefully at your work yet realise...but something that might help all of you is a quick and dirty heuristic for why we should expect convergence:
What is the probability of an arbitrary string of length m not containing n consecutive heads somewhere?
Well let's call strings containing n consecutive heads good and let's count those that are bad.
We do this crudely by breaking the m character string into "blocks" of n. There are roughly m/n of these.
Every bad string cannot have any block comprised entirely of H's. So the probability of an arbitrary m-string being bad is roughly bounded by (1-1/2^n)^(m/n), which clearly tends to zero as m->inf.
This also tells us that the average should be finite (which isn't obvious), because the decay in m is exponential and thus beats the linear factor m that occurs in the sum for the average.
What is the probability of an arbitrary string of length m not containing n consecutive heads somewhere?
Well let's call strings containing n consecutive heads good and let's count those that are bad.
We do this crudely by breaking the m character string into "blocks" of n. There are roughly m/n of these.
Every bad string cannot have any block comprised entirely of H's. So the probability of an arbitrary m-string being bad is roughly bounded by (1-1/2^n)^(m/n), which clearly tends to zero as m->inf.
This also tells us that the average should be finite (which isn't obvious), because the decay in m is exponential and thus beats the linear factor m that occurs in the sum for the average.
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anomalousdecay
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Its working fine on my end.
Though yesterday between 11pm-2am I had issues with it rendering.
EDIT: Its not working for me too either at the moment.
Did this happen to you:
http://community.boredofstudies.org...319052/problems-latex-mode-2.html#post6551707
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seanieg89
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I think that's what I got
.post the solution pls haha
Squar3root
realest nigga
yeah post it lol
seanieg89
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View attachment avg number of tosses.pdf
I apologise for the sloppy texing. Hopefully it makes sense, but please point out anything that seems incorrect as typos are possible and the solution is my own.
I apologise for the sloppy texing. Hopefully it makes sense, but please point out anything that seems incorrect as typos are possible and the solution is my own.
seanieg89
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Haven't thought about this sort of stuff (and in particular this problem) for ages. I suspect it would be quite hard if not impossible to get a closed form for the pdf. Otherwise summing a series involving the pdf would be the natural way to find the mean.
Given that the exploitation of symmetry was the only was I could originally find for computing the mean, this doesn't bode well for closed form computations.
(It is entirely possible I missed something, this is not an area of strength for me.)